题目描述
Bessie likes downloading games to play on her cell phone, even though she doesfind the small touch screen rather cumbersome to use with her large hooves.
She is particularly intrigued by the current game she is playing.The game starts with a sequence of NN positive integers (2 \leq N\leq 2482≤N≤248), each in the range 1 \ldots 401…40. In one move, Bessie cantake two adjacent numbers with equal values and replace them a singlenumber of value one greater (e.g., she might replace two adjacent 7swith an 8). The goal is to maximize the value of the largest numberpresent in the sequence at the end of the game. Please help Bessiescore as highly as possible!
给定一个1*n的地图，在里面玩2048，每次可以合并相邻两个（数值范围1-40），问最大能合出多少。注意合并后的数值并非加倍而是+1，例如2与2合并后的数值为3。
输入格式
The first line of input contains NN, and the next NN lines give the sequence
of NN numbers at the start of the game.
输出格式
Please output the largest integer Bessie can generate.
输入输出样例
输入 #1复制
4 1 1 1 2
输出 #1复制
3
说明/提示
In this example shown here, Bessie first merges the second and third 1s to
obtain the sequence 1 2 2, and then she merges the 2s into a 3. Note that it is
not optimal to join the first two 1s.
f[i][j]表示将序列中的第i个数合并到第j个数全部合并所能得到的最大数值
转移条件：f[l][k]==f[k+1][r]
同时注意由于f[1][n]不一定能全部合并，所以最大值是在合并的过程中用一个变量不停维护最大值
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=500;
typedef long long LL;
int a[maxn];
int f[maxn][maxn];//f[i][j]表示将序列中的第i个数合并到第j个数全部合并所能得到的最大数值
int main(void)
{cin.tie(0);std::ios::sync_with_stdio(false);int n;cin>>n;for(int i=1;i<=n;i++) cin>>a[i];for(int i=1;i<=n;i++) f[i][i]=a[i];int ans=-1;for(int len=2;len<=n;len++)for(int l=1;l+len-1<=n;l++){int r=l+len-1;for(int k=l;k<r;k++){if(f[l][k]==f[k+1][r]){f[l][r]=max(f[l][r],f[l][k]+1); } ans=max(ans,f[l][r]);//找出某个区间内合并的最大 } }
// for(LL l=1;l<=n;l++)
// for(LL r=l;r<=n;r++)
// {
// cout<<"f["<<l<<"]["<<r<<"]="<<f[l][r]<<endl;
// }cout<<ans<<endl;
return 0;
}